Problem: The grades on a physics midterm at Almond are normally distributed with $\mu = 75$ and $\sigma = 4.0$. Umaima earned a $79$ on the exam. Find the z-score for Umaima's exam grade. Round to two decimal places.
Solution: A z-score is defined as the number of standard deviations a specific point is away from the mean We can calculate the z-score for Umaima's exam grade by subtracting the mean $(\mu)$ from her grade and then dividing by the standard deviation $(\sigma)$ $ { z = \dfrac{x - {\mu}}{{\sigma}}} $ $ { z = \dfrac{79 - {75}}{{4.0}}} $ ${ z \approx 1.00}$ The z-score is $1.00$. In other words, Umaima's score was $1.00$ standard deviation above the mean.